If $x^2 + \dfrac{1}{4x^2} = 8, \text{ find } x^3 + \dfrac{1}{8x^3}$.

We know that,

$\Rightarrow \Big(x + \dfrac{1}{2x}\Big)^2 = x^2 + \dfrac{1}{4x^2} + 2 \times x \times \dfrac{1}{2x} \\[1em] \Rightarrow \Big(x + \dfrac{1}{2x}\Big)^2 = x^2 + \dfrac{1}{4x^2} + 1 \\[1em] \therefore x^2 + \dfrac{1}{4x^2} = \Big(x + \dfrac{1}{2x}\Big)^2 - 1$

Substituting values in above equation we get,

$\Rightarrow 8 = \Big(x + \dfrac{1}{2x}\Big)^2 - 1 \\[1em] \Rightarrow \Big(x + \dfrac{1}{2x}\Big)^2 = 8 + 1 \\[1em] \Rightarrow \Big(x + \dfrac{1}{2x}\Big)^2 = 9 \\[1em] \Rightarrow x + \dfrac{1}{2x} = \sqrt{9} \\[1em] \Rightarrow x + \dfrac{1}{2x} = \pm 3.$

We know that,

$\Rightarrow (a + b)^3 = a^3 + b^3 + 3ab(a + b) \\[1em] \Rightarrow \Big(x + \dfrac{1}{2x}\Big)^3 = x^3 + \Big(\dfrac{1}{2x}\Big)^3 + 3 \times x \times \dfrac{1}{2x}\Big(x + \dfrac{1}{2x}\Big) \\[1em] \Rightarrow \Big(x + \dfrac{1}{2x}\Big)^3 = x^3 + \dfrac{1}{8x^3} + \dfrac{3}{2}\Big(x + \dfrac{1}{2x}\Big) \\[1em] \therefore x^3 + \dfrac{1}{8x^3} = \Big(x + \dfrac{1}{2x}\Big)^3 - \dfrac{3}{2}\Big(x + \dfrac{1}{2x}\Big).$.

Case 1: $x + \dfrac{1}{2x} = 3$ substituting values we get,

$x^3 + \dfrac{1}{8x^3} = 3^3 - \dfrac{3}{2} \times 3 \\[1em] = 27 - \dfrac{9}{2} \\[1em] = \dfrac{54 - 9}{2} \\[1em] = \dfrac{45}{2}.$

Case 2 : $x + \dfrac{1}{2x} = -3$, substituting values we get,

$x^3 + \dfrac{1}{8x^3} = (-3)^3 - \dfrac{3}{2} \times (-3) \\[1em] = -27 + \dfrac{9}{2} \\[1em] = \dfrac{-54 + 9}{2} \\[1em] = -\dfrac{45}{2}.$

Hence, $x^3 + \dfrac{1}{8x^3} = \pm \dfrac{45}{2} = \pm 22\dfrac{1}{2}$.